120/208VAC panel main breaker size
120/208VAC panel main breaker size
(OP)
Hello,
I have a 120/208 Vac panel that is fully populated and due to the nature of this panel installation there is not a need to provide additional expansion. The load is a mix of 120 and 208 lighting loads and receptacle loads (no motors etc.)
The total load on the panel is 26,688VA.
My question is: what is the minimum size main breaker that I can use to protect the panel.
I think it should be (26688VA/(208*1.73))*1.25= 92amps. So I should use a 100amp breaker.
Is this correct??
I have a 120/208 Vac panel that is fully populated and due to the nature of this panel installation there is not a need to provide additional expansion. The load is a mix of 120 and 208 lighting loads and receptacle loads (no motors etc.)
The total load on the panel is 26,688VA.
My question is: what is the minimum size main breaker that I can use to protect the panel.
I think it should be (26688VA/(208*1.73))*1.25= 92amps. So I should use a 100amp breaker.
Is this correct??






RE: 120/208VAC panel main breaker size
RE: 120/208VAC panel main breaker size
RE: 120/208VAC panel main breaker size
Like rbulsara said, match the panelboard rating.
RE: 120/208VAC panel main breaker size
The problem is that I am unsure of the calculation because the loads are both 208 and 120.
RE: 120/208VAC panel main breaker size
This type of sizing always leads to regrets and bad mouthing of the person that was penny wise and pound foolish to install it in the first place.
RE: 120/208VAC panel main breaker size
BUT, If you insist;
Step one.> Start with your 208 volt loading. It is a waste of time to try to calculate the exact current in each phase. If you do, so what? You can't get a breaker with one pole 80 amps the second pole 100amps and the third pole 93 amps.
Assume that the 208 volt loads are balanced but use the heaviest phase to phase load for all three phases.
Example; Phase "A" to "B" 37 Amps, Phase "B" to "C" 30 Amps, and Phase "C" to "A" 40 Amps.
Assume that all three phases are loaded to 40 Amps for your calculations.
Step 2.> 40 Amps plus 40 Amps = 80 Amps x .866 = 69 Amps.
This (69 Amps) is your phase current from the 208 volt loads and may now be added directly to the 120 volt loads.
Add the heaviest single phase load. This is your absolute minimum loading. Check your code to see if the feeders must be over sized and check the maximum temperature rating that may be used when determining the feeder ampacity.
Now I expect to hear about a year or so from now;
"Help! Our panel worked well for a year but a component in one of the loads failed and the replacement draws more current, the main breaker..............."
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: 120/208VAC panel main breaker size
In step 2, I am not clear on why you are adding 40 and 40?
and where does the factor .866 come from?
Could I not just take the sum of the VAs for the heaviest loaded phase and divide it by 120?
RE: 120/208VAC panel main breaker size
Add the heaviest single phase load.
This would be better if I had written "Add the line to neutral current from the heaviest loaded phase."
How about if I start over.
Start with the balanced three phase loads.
Add 1.73 of the unbalanced current from the heaviest loaded line to line connection. (.866 is 1/2 of 1.73)
Add the line to neutral current from the phase with the heaviest line to neutral current.
OR
Consider 13 identical loads each drawing 14 amps single phase at 208 volts.
We can divide these loads into a balanced load with 4 on each phase and 1 temporarily left over.
Each group of 4 will draw 4 x 14 = 56 amps.
But when we combine the 56 amps from A-B to the 56 amps from B-C, we get 1.73 X 56 amps = 97 amps rather than 112 amps.
The resultant current is now at the correct phase angle to be added directly to the line to neutral currents.
Now take the left over load and assume that there are 2 other identical loads and that it is balanced. Multiply by 1.73 to correct the phase angle and add this current to the others. With only one odd load the resultant current is a little high but with two odd loads this will be the current on the highest loaded phase.
The reason that you can't just add the VA is the phase angle difference between line to line loads and line to neutral loads.
Note, with only one
Bill
--------------------
"Why not the best?"
Jimmy Carter